Animated Solution for Chemistry - Chemical Kinetics: For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is ........... (Integer answer)

Welcome to the fascinating world of chemical kinetics! Today, we are tackling a classic problem involving first-order reactions. Imagine a radioactive sample or a chemical decomposing over time. For a first-order reaction, the concentration doesn't just drop linearly; it drops exponentially.

To find the exact time taken for any percentage of the reaction to complete, we rely on our trusty integrated rate law. This master equation beautifully connects the rate constant $k$, the time $t$, and the initial and final concentrations. It's the ultimate tool for these kinds of problems.

Let's set up the equation for $50\%$ completion, which is famously known as the half-life ($t_{1/2}$). Here, exactly half of our initial reactant is consumed. So, if we start with $100$ units, the remaining concentration is $50$ units. Plugging these values into our formula gives us:

Similarly, let's look at $75\%$ completion. This means three-quarters of our reactant has vanished into products! Only $25\%$ remains in the flask. We substitute $100$ for the initial amount and $25$ for the remaining amount into our rate law:

Now, here is the crucial logical step. Since the rate constant $k$ is an intrinsic property of the reaction at a given temperature, it remains perfectly constant. Therefore, we can confidently equate the two expressions we just derived. Notice how those bulky $2.303$ terms are just waiting to be cancelled out.

Let's simplify the math. On the left, we have $\log (2)$. On the right, we have $\log (4)$. But wait, $4$ is just $2^2$! Using the power rule of logarithms, $\log (2^2)$ becomes $2\log (2)$. Now, the $\log (2)$ terms cancel out beautifully from both sides.

After that satisfying cancellation, we are left with a remarkably simple relation:

Rearranging this gives us our final answer:

The time required for $75\%$ completion is exactly twice the time required for $50\%$ completion.

The Intuitive Shortcut: You can even solve this intuitively without touching a single logarithm! One half-life leaves $50\%$ of the reactant. A second half-life halves that remaining $50\%$, leaving exactly $25\%$. If $25\%$ is left, it means $75\%$ has reacted! Therefore, it takes exactly two half-lives to reach $75\%$ completion. Keep these mental shortcuts handy for your exams!